What is the tangent of (7-6x)^{1/2},\at x=-1 ?

The tangent of (7-6x)^{1/2},\at x=-1 is y=-3/(13^{1/2)}x+13^{1/2}-3/(13^{1/2)}

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tangent of (7-6x)^{1/2},\at x=-1

tangent of

(7 - 6 x)^{\frac{1}{2}}, at x = - 1

y = - \frac{3}{1 3 ^{\frac{1}{2}}} x + 1 3^{\frac{1}{2}} - \frac{3}{1 3 ^{\frac{1}{2}}}

Solution steps

Find the tangent point:

(- 1, 1 3^{\frac{1}{2}})

Find the slope of

y = (7 - 6 x)^{\frac{1}{2}} : \frac{d y}{d x} = - \frac{3}{( 7 - 6 x ) ^{\frac{1}{2}}}

Compute the slope of

y = (7 - 6 x)^{\frac{1}{2}}

x = - 1 : m = - \frac{3}{1 3 ^{\frac{1}{2}}}

Find the line with slope m=

- \frac{3}{1 3 ^{\frac{1}{2}}}

and passing through

(- 1, 1 3^{\frac{1}{2}}) : y = - \frac{3}{1 3 ^{\frac{1}{2}}} x + 1 3^{\frac{1}{2}} - \frac{3}{1 3 ^{\frac{1}{2}}}

y = - \frac{3}{1 3 ^{\frac{1}{2}}} x + 1 3^{\frac{1}{2}} - \frac{3}{1 3 ^{\frac{1}{2}}}

\int_{1}^{2} \frac{1}{x ^{3}} d x

f (x) = x^{2} + x, (2, 6)

\frac{d}{d x} (cos^{3} (x) + sin^{3} (x))

x \to \infty lim (\frac{3}{2})

\int_{0}^{π} 2 π x (sin (x)) d x

What is the tangent of (7-6x)^{1/2},\at x=-1 ?
The tangent of (7-6x)^{1/2},\at x=-1 is y=-3/(13^{1/2)}x+13^{1/2}-3/(13^{1/2)}